chapter 6 review mth 065 – elementary algebra. the graph of f(x) = x 2 + bx + c vs. solutions of x...
TRANSCRIPT
The Graph of f(x) = x2 + bx + cvs.
Solutions of x2 + bx + c = 0vs.
Factorization of x2 + bx + c
x
y
The graph is a parabola.
The solutions are the x-intercepts.
The factors are (x – m)(x – n), where m and n are the x-intercepts.
The Graph of f(x) = x2 + bx + cvs.
Solutions of x2 + bx + c = 0vs.
Factorization of x2 + bx + cThree possibilities …
x = 3, –1
(x – 3)(x + 1)
x = –2
(x + 2)2
No Solution
Does not Factor
What if an x-intercept is the origin?
The Principle of Zero ProductsIf ab = 0, then either a = 0, b = 0, or both.
Using this Principle
Solve: x(x – 5)(2x + 1) = 0
Solution:x = 0x – 5 = 0 x = 52x + 1 = 0 x = –½
Therefore …
x = 0, 5, –½
Factoring – Step 1 … always!
Factor out common factors.Examples:
Constant: 6x2 – 12x – 21 = 3(2x2 – 4x – 7)
Variable:x3 + 5x2 – 2x = x(x2 + 5x – 2)
Both: –4x4 + 32x3 – 20x2 = –4x2(x2 – 8x + 5)
Factoring by Grouping (4 terms)
x3 – 2x2 + 5x – 10
Make two groups of two terms.
(x3 – 2x2) + (5x – 10)
Factor out common factors from each group.
x2(x – 2) + 5(x – 2)
If the remaining binomials are identical, factor them out.
(x – 2)(x2 + 5)
NOTE: Be careful with the sign of the third term when it is negative!
Factoring: x2 + bx + cLeading coefficient = 1Find two numbers m & n where …
mn = c m + n = b
Then the factors are … (x + m)(x + n)
Signs …If c is positive, then both numbers have the
same sign as b.If c is negative, then the “larger” one has the
same sign as b and the other one has the opposite sign.
Solving: x2 + bx + c = 0
All terms to the left or right (one side MUST be 0).
Leading coefficient = 1
Factor the polynomial …
(x + m)(x + n) = 0
Set each factor equal to 0 and solve.
i.e. Solutions are: x = – m, –n
Factoring: ax2 + bx + cLeading coefficient > 1Find two numbers m & n where …
mn = ac m + n = b
Rewrite the polynomial: ax2 + mx + nx + c
Factor by grouping (be careful when nx is negative).
Signs …If ac is positive, then both numbers have the same
sign as b.If ac is negative, then the “larger” one has the
same sign as b and the other one has the opposite sign.
Solving: ax2 + bx + c = 0
All terms to the left or right (one side MUST be 0).
Leading coefficient > 1
Factor the polynomial (ac method) …
(rx + s)(tx + u) = 0
Set each factor equal to 0 and solve.
i.e. Solutions are: x = – s/r, –u/t
Leading Coefficient Negative
Factor out –1 first, and then proceed as before.
Example: –8x2 + 10x + 3 = –[8x2 – 10x – 3] = –[8x2 – 12x + 2x – 3] = –[(8x2 – 12x) + (2x – 3)] = –[4x(2x – 3) + 1(2x – 3)] = –(2x – 3)(4x + 1)
Factoring: Special Case #1
Perfect Square Polynomials
ax2 + bx + cIf a is a perfect square … a = m2
If c is a perfect square … c = n2
If b = 2mn
Then it factors as …If b is positive: (mx + n)2
If b is negative: (mx – n)2
Factoring: Special Case #2
Difference of two squares.
ax2 – cIf a is a perfect square … a = m2
If c is a perfect square … c = n2
Then it factors as …(mx + n)(mx – n)
NOTE: It must be subtraction! If it is addition, it will not factor.
Factoring Polynomials in Quadratic Form
af(x) 2 + bf(x) + c
Let y = f(x) (i.e. replace f(x) with y)
ay2 + by + c
Factor …
(my + n)(ry + s)
Substitute back (i.e. replace y with f(x)
(mf(x) + n)(rf(x) + s)
Simplify … if possible.